Question 768: Centre of Mass and Rotational Motion

Question

The figure shows a system consisting of (i) a ring of outer radius 3R rolling clockwise without
slipping on a horizontal surface with angular speed $$\omega $$ and (ii) an inner disc of radius 2R rotating
anti-clockwise with angular speed $${\omega  \over 2}$$. The ring and disc are separated by frictionless ball
bearings. The point P on the inner disc is at a distance R from the origin, where OP makes an
angle of $$30^\circ $$ with the horizontal. Then with respect to the horizontal surface,

Accepted Answer

The outer ring rolls without slipping, meaning the point on the ring in contact with the ground (denoted as point C) is stationary, i.e., $ \vec{v}_{\mathrm{C}} = \overrightarrow{0} $. Therefore, the velocity of point O is :
$$  \vec{v}_{\mathrm{O}} = \vec{v}_{\mathrm{C}} + \vec{\omega}_{\mathrm{o}} 	imes \vec{r}_{\mathrm{CO}} = \overrightarrow{0} + \omega \hat{\jmath} 	imes 3R \hat{k} = 3R\omega \hat{\imath}  $$
Point P is located on the inner disc, which has an angular velocity $ \vec{\omega}_{\mathrm{i}} = -\omega / 2 \hat{\jmath} $. The position vector from O to P is given by :
$$  \vec{r}_{\mathrm{OP}} = R \cos 30^{\circ} \hat{\imath} + R \sin 30^{\circ} \hat{k} = \frac{\sqrt{3} R}{2} \hat{\imath} + \frac{R}{2} \hat{k}  $$
Thus, the velocity of point P is calculated as follows :
$$  \begin{aligned}  \vec{v}_{\mathrm{P}} & = \vec{v}_{\mathrm{O}} + \vec{\omega}_{\mathrm{i}} 	imes \vec{r}_{\mathrm{OP}} \  & = 3R\omega \hat{\imath} + \left(-\frac{\omega}{2} \hat{\jmath}ight) 	imes \left(\frac{\sqrt{3} R}{2} \hat{\imath} + \frac{R}{2} \hat{k}ight) \  & = \frac{11\omega R}{4} \hat{\imath} + \frac{\sqrt{3} \omega R}{4} \hat{k} .  \end{aligned}  $$

Answer

the point O has linear velocity $$3R\omega \widehat i$$

Answer

the point P has linear velocity $${{11} \over 4}R\omega \widehat i + {{\sqrt 3 } \over 4}R\omega \widehat k$$

Answer

the point P has linear velocity $${{13} \over 4}R\omega \widehat i - {{\sqrt 3 } \over 4}R\omega \widehat k$$

Answer

the point P has linear velocity $$\left( {3 - {{\sqrt 3 } \over 4}} \right)R\omega \widehat i + {1 \over 4}R\omega \widehat k$$

Centre of Mass and Rotational Motion

Choose the correct answer: